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DefinitionDefinition: AI-adaptedProof: Not applicableprecheck passjudge pass (gpt-6.1-sol)
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eta-orthogonality and orthogonal complements in a Lorentz vector space

Definition

Let (V,η) be a Lorentz vector space of signature (−,+,+,+) (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space). Two vectors u,v∈V are η-orthogonal, written u⊥v, when η(u,v)=0. For a subset S⊆V the orthogonal complement of S is

S⊥:={v∈V:η(v,s)=0 for every s∈S},

and for a vector u one writes u⊥:={u}⊥.

Elementary properties. For every subset S⊆V:

  • S⊥ is a linear subspace of V: if v,w∈S⊥ and λ∈R, then for every s∈S bilinearity gives η(v+λw,s)=η(v,s)+λη(w,s)=0, so v+λw∈S⊥. In particular the zero vector lies in S⊥ and S⊥ is never empty.
  • S⊆(S⊥)⊥: if s∈S, then η(v,s)=0 for every v∈S⊥ by the definition of S⊥, and symmetry of η gives η(s,v)=0 for every v∈S⊥, that is, s∈(S⊥)⊥.
  • S⊥=(span⁡S)⊥: the inclusion S⊆span⁡S gives (span⁡S)⊥⊆S⊥; conversely, if v∈S⊥ and u=∑iλisi is a finite linear combination with si∈S, then linearity in the second argument gives η(v,u)=∑iλiη(v,si)=0, so v∈(span⁡S)⊥.

In particular, for every u∈V one has u∈u⊥ exactly when η(u,u)=0, that is, exactly when u is null or u=0 (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

A nonzero vector can be orthogonal to itself. Take a basis e0,e1,e2,e3 of V in which the matrix of η is diagonal with diagonal entries −1,1,1,1; such a basis exists by clause 2 of the definition of a Lorentz form. Then v:=e0+e1 is nonzero while η(v,v)=η(e0,e0)+2η(e0,e1)+η(e1,e1)=−1+0+1=0, so v∈v⊥ with v≠0. This is exactly what fails for a positive-definite inner product and is why orthogonal complements in a Lorentz vector space need not be positive definite.

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